\(E=k\dfrac{q}{r^2}=9.10^9\dfrac{10^{-8}}{\left(20.10^{-2}\right)^2}=2250\left(\dfrac{V}{m}\right)\)
Khi \(E_1=10^5\left(\dfrac{V}{m}\right)\) thì khoảng cách là: \(r^2=k\dfrac{q}{E_1}=9.10^9\dfrac{10^{-8}}{10^5}\Rightarrow r=0,03\left(m\right)=3\left(cm\right)\)